Question 442318
Recall that vertex form is y=(x-h)^2 + k. Where (h,k) is your vertex.

Also recall that given standard form Ax^2 - Bx + C, the vertex is (-B/2A, f(-B/2A)).

Well x^2-4x-4 is pretty standard.  Thus the x-coordinate of the vertex is -(-4)/2.
OR 4/2 ... which is 2.

now what is f(2)?  2^2 - 8 - 4 = -8...  So our y-coordinate is -8.

The vertex is (2,-8) OR  h=2  and k=-8.

So to put this in vertex form, we simply substitute.

for y=(x-h)^2 + k and h=2 k =-8 we get:

{{{y = (x-2)^2 - 8}}}

Hope this helped!